Differential Equations and Linear Algebra
2250-4 [12:55 class] at 1:00pm on 3 May 2010
Instructions. The time allowed is 120 minutes. The examination consists of eight problems, one for each
of chapters 3, 4, 5, 6, 7, 8, 9, 10, each problem with multiple parts. A chapter represents 15 minutes on the
final exam.
Each problem on the final exam represents several textbook problems numbered (a), (b), (c), · · ·. Each chapter
(3 to 10) adds at most 100 towards the maximum final exam score of 800. The final exam grade is reported
as a percentage 0 to 100, as follows:
Final Exam Grade =
Sum of scores on eight chapters
.
8
• Calculators, books, notes and computers are not allowed.
• Details count. Less than full credit is earned for an answer only, when details were expected. Generally,
answers count only 25% towards the problem credit.
• Completely blank pages count 40% or less, at the whim of the grader.
• Answer checks are not expected and they are not required. First drafts are expected, not complete
presentations.
• Please prepare exactly one stapled package of all eight chapters, organized by chapter. All scratch
work for a chapter must appear in order. Any work stapled out of order could be missed, due to multiple
graders.
• The graded exams will be in a box outside 113 JWB; you will pick up one stapled package.
• Records will be posted at the Registrar’s web site on WEBct. Recording errors are reported by email.
Final Grade. The final exam counts as two midterm exams. For example, if exam scores earned were 90,
91, 92 and the final exam score is 89, then the exam average for the course is
Exam Average =
90 + 91 + 92 + 89 + 89
= 90.2.
5
Dailies count 30% of the final grade. The course average is computed from the formula
Course Average =
70
30
(Exam Average) +
(Dailies Average).
100
100
Please discard this page or keep it for your records.
Name
12:55pm Version 1
Differential Equations and Linear Algebra 2250-4 [12:55 class]Final Exam at 1:00pm on 3 May 2010
Ch3. (Linear Systems and Matrices) Complete all problems.
[10%] Ch3(a): Check the correct box. Incorrect answers lose all credit.
Part 1. [5%]:
True or
False:
If the 10 × 10 matrices A and B are lower triangular, then AB is triangular.
Part 2. [5%]:
True or
False:
If a 3 × 3 matrix A has a row of zeros, then for all vectors b, the equation Ax = b has infinitely many
solutions x.
Answer: True. False.
[30%] Ch3(b): Determine which values of k correspond to (1) a unique solution, (2) infinitely many
solutions and (3) no solution, for the system Ax = b given by


0 k−2 k−3


4
k ,
A= 1
1
4
3


−1


b =  1 .
k

Answer:
1
Elimination methods with swap, combo, multiply give  0
0
4
k
k−2
0
0
3−k

1
k − 2 .
k−1
Then (1) Unique solution for k 6= 2 and k 6= 3; (2) No solution for k = 3 [signal equation]; (3)
Infinitely many solutions for k = 2.
[20%] Ch3(c): Let matrix C and vector b be defined by the equations


−2
3
0


4 ,
C =  0 −2
1
0 −3


1


b =  2 .
3
Let I denote the 3×3 identity matrix. Find the value of x2 by Cramer’s Rule in the system (3I +C)x = b.
Answer: x2 = ∆2 /∆, ∆2 = −8, ∆ = det(3I + C) = 12, x2 = −2/3.
[20%]Ch3(d): Display
 the entry in row 3, column 4 of the adjugate matrix [or adjoint matrix] of
0 2 −1 0
 0 0
4 1 


A=
.
 1 3 −2 0 
0 1
1 0
Answer: answer = cofactor of A in row 4, column 3 = (−1)7 times minor of A in 4,3 = −2.
A−1
=
4 6
0 2
!
[20%] Ch3(e): Assume
Answer: A =
2 −6
0
4
, AT =
!
. Find the transpose of A.
4 0
6 2
!
.
Scores
Ch3.
Ch4.
Ch5.
Ch6.
Ch7.
Ch8.
Ch9.
Ch10.
Staple this page to the top of all Ch3 work.
Name
12:55pm Version 1
Differential Equations and Linear Algebra 2250-4 [12:55 class]Final Exam at 1:00pm on 3 May 2010
Ch4. (Vector Spaces) Complete all problems.
[20%] Ch4(a): Define S to be the set of all vectors x in R4 such that x1 + x3 = 0 and x3 + x4 = x1 .
Prove that S is a subspace of R4 .

1 0 1
 −1 0 1
Answer: Let A = 

 0 0 0
0 0 0
Apply the kernel theorem. This is

0
1 

. Then the restriction equations can be written as Ax = 0.
0 
0
theorem 2 in section 4.2 of Edwards-Penney.
[20%] Ch4(b): Give an example of three vectors v1 , v2 , v3 for which the nullity of their augmented
matrix is two.
Answer: Let v1 = v2 = v3 for any nonzero vector v1 .
[30%] Ch4(c): Apply an independence test to the vectors below. Report independent or dependent.
Details count.






0
3
−1
 −1 
 0 
 1 






v1 = 
.
 , v3 = 
 , v2 = 
 −1 
 1 
 2 
0
0
0
Answer: Independent. The rank of the augmented matrix of the three vectors is 3.
[30%] Ch4(d): Find a basis of fixed vectors in R4 for the solution space of Ax = 0, where the 4 × 4
matrix A is given below.


3 −1
1 1
 1 −1
1 0 


A=
.
 2
0
0 1 
1
1 −1 1

Answer:
1 0

 0 1

rref (A) = 

 0 0

0 0
0
1/2



0
 1
−1 1/2 

, basis = 


 1
0
0 

0
0
0
Staple this page to the top of all Ch4 work.
 


,





−1/2
−1/2
0
1





Name
12:55pm Version 1
Differential Equations and Linear Algebra 2250-4 [12:55 class]Final Exam at 1:00pm on 3 May 2010
Ch5. (Linear Equations of Higher Order) Complete all problems.
[10%] Ch5(a): Report the general solution y(x) of the differential equation
3
d3 y
d2 y
dy
+
10
+3
= 0.
3
2
dx
dx
dx
Answer: Characteristic equation r(3r + 1)(r + 3) = 0 has roots 0, −1/3, −3. The general
solution y(x) is a linear combination of the atoms 1, e−x/3 , e−3x .
[20%] Ch5(b): Given a damped spring-mass system mx00 (t) + cx0 (t) + kx(t) = 0 with m = 2, c = 2 + a,
k = 1 + a and a > 0 a symbol, calculate all values of symbol a such that the solution x(t) is overdamped. Please, do not solve the differential equation!
Answer: The discriminant must be positive. Then (2 + a)2 − 8(1 + a) > 0 or a2 − 4a − 4 > 0.
[20%] Ch5(c): A particular solution of the differential equation x00 + 2x0 + 17x = 50 cos(3t) is
x(t) = 4 cos 3t + 12e−t sin 4t + 3 sin 3t + 15e−t cos 4t.
Identify the steady-state solution xss (t).
Answer: The transient solution is the sum of all terms with limit zero. The steady-state is the
sum of the remaining terms. Then xss = 4 cos 3t + 3 sin 3t.
[20%] Ch5(d): Determine a basis of solutions of a homogeneous constant-coefficient linear differential
equation, given it has characteristic equation
r(r2 + r)2 ((r + 1)2 + 7)2 = 0.
√
√
Answer: The roots are 0, 0, 0, −1, −1, −1± 7i, −1±
7i. By Euler’s
a√
basis is the set √
√
√ theorem,
2 −x
−x −x
−x
−x
−x
of atoms for these roots: 1, x, x , e
, xe
,e
cos( 7x), e
sin( 7x), xe
cos( 7x), xe
sin( 7x).
[30%] Ch5(e): Determine the shortest trial solution for yp according to the method of undetermined
coefficients. Do not evaluate the undetermined coefficients!
d4 y
d2 y
+
9
= 2x2 + 3x sin 3x + 4ex
dx4
dx2
Answer: Let f (x) = 2x2 + 3x sin 3x + 4ex . The atoms in f are x2 , x sin 3x, ex . The completed
set of atoms, including all lower-power related atoms, is 1, x, x2 , cos 3x, x cos 3x, sin 3x, x sin 3x, ex .
There are 8 atoms in this list. A theorem says that the shortest trial solution contains 8 atoms.
Break the 8 atoms into four groups, each with the same base atom: group 1 == 1, x, x2 ; group 2
== cos 3x, x cos 3x; group 3 == sin 3x, x sin 3x; group 4 == ex . Modify each group, to remove
any solution of the homogeneous equation y (4) + 9y (2) = 0. Then the four groups are replaced by
group 1∗ == x2 , x3 , x4 ; group 2∗ == x cos 3x, x2 cos 3x; group 3∗ == x sin 3x, x2 sin 3x; group
4∗ == ex . The shortest trial solution is a linear combination of these last eight atoms.
Staple this page to the top of all Ch5 work.
Name
12:55pm Version 1
Differential Equations and Linear Algebra 2250-4 [12:55 class]Final Exam at 1:00pm on 3 May 2010
Ch6. (Eigenvalues and Eigenvectors) Complete all problems.




[40%] Ch6(a): Find the eigenvalues of the matrix A = 


0 −2 −5
0 0
3
0 −12
3 0
0
0
1 −1 0
0
0
1
3 0
0
0
5
1 3




.


To save time, do not find eigenvectors!
2
2
Answer: The
√ characteristic polynomial is det(A−rI) = (r +6)(3−r)(r −2) . The eigenvalues
are 2, 2, 3, ± 6i. Determinant expansion is by the cofactor method along column 5. This reduces
it to a 4 × 4 determinant, which can be expanded as a product of two quadratics.
[30%] Ch6(b): Find
the eigenvectors corresponding to complex eigenvalues −1 ± 3i for the 2 × 2 matrix
!
−1
3
A=
.
−3 −1
Answer:
−1 + 3i,
[30%] Ch6(c): Let A =
−i
1
!!
−7 4
−12 7
1,
−1 − 3i,
,
i
1
!!
!
. Circle possible eigenpairs of A.
1
2
!!
,
2,
2
1
!!
,
−1,
2
3
!!
.
Answer: The first and the last, because the test Ax = λx passes in both cases.
Staple this page to the top of all Ch6 work.
Name
12:55pm Version 1
Differential Equations and Linear Algebra 2250-4 [12:55 class]Final Exam at 1:00pm on 3 May 2010
Ch7. (Linear Systems of Differential Equations) Complete all problems.
[30%] Ch7(a): Solve for the general solution x(t), y(t) in the system below. Use any method that
applies, from the lectures or any chapter of the textbook.
dx
dt
dy
dt
= x + y,
= 6x + 2y.
!
1 1
Answer: Define A =
. The eigenvalues −1, 4 are roots of the characteristic equation
6 2
det(A − rI) = (r + 1)(r − 4) = 0. By Cayley-Hamilton-Zeibur, x(t) = c1 e−t + c2 e4t . Using the
first differential equation x0 = x + y implies y(t) = x0 − x = −2c1 e−t + 3c2 e4t .
[30%] Ch7(b): Let A be an n × n matrix of real numbers. State three different methods for solving the
system ~u0 = A~u, which you learned in this course.
Answer: Eigenanalysis, Zeibur-Cayley-Hamilton, Laplace resolvent, exponential matrix.
[40%] Ch7(c): Define


−3
4 −10


2
0 
A= 0
5 −4
12
The eigenvalues of A are 7, 2, 2. Apply the eigenanalysis method, which requires eigenvalues and
eigenvectors, to solve the differential system u0 = Au.


1
−10
4 −10
0 −5
0 . Then rref (A1 ) =  0
Form A1 = A − (7)I = 
0
5 −4
5

Answer:
0
1
0

1
0 . The
0
last frame algorithm implies general solution of A1 x = 
0 is x1= −t1 , x2 = 0, x3 = t1 . Take
the partial on symbol t1 to obtain the eigenvector v1 = 

−5
 0
5


4 −10
1 − 45


0
0 . Then rref (A2 ) =
0
0
−4
10
0
0
−1
0 . Repeat with A2 = A − (2)I =
1

2
0 . The last frame algorithm implies general
0
solution of A1 x = 0 is x1 = 4t1 /5− 2t
t1 , x3 = t2 . Take the partial on symbols t1 ,
2 , x2 = 
4
5
t2 to obtain the eigenvectors v2 =  1 , v3 = 
0
u(t) = c1 e7t v1 + c2 e2t v2 + e2t v3 .
Staple this page to the top of all Ch7 work.
−2
0 . The general solution of u0 = Au is
1
Name
12:55pm Version 1
Differential Equations and Linear Algebra 2250-4 [12:55 class]Final Exam at 1:00pm on 3 May 2010
Ch8. (Matrix Exponential) Complete all problems.
[30%] Ch8(a): Consider the 2 × 2 system
x0 = x,
y 0 = −2y,
x(0) = 1, y(0) = 2.
Solve the system u0 = Au for u, using the matrix exponential eAt .
Answer:
x(t)
y(t)
!
1
2
= eAt
!
et
0
−2t
0 e
=
!
1
2
!
et
=
2e−2t
!
.
[40%] Ch8(b): Display the matrix form of variation of parameters for the 2 × 2 system. Then integrate
to find one particular solution.
x0 = x + 3,
y 0 = −2y + 1.
Answer:
e−As
=
R
up (t) = eAt 0t e−As
diag(e−s , e2s ),
Finally, up (t) =
and
Rt
0
diag(et , e−2t )
2
1
!
ds. Then eAt = diag(et , e−2t ) from the first problem,
3
1
diag(e−s , e2s )
3 − 3e−t
1 2t
1
2e − 2
!
=
!
ds =
Rt
0
diag(3e−s , e2s )ds
−3 + 3et
1 −2t
1
2 − 2e
=
3 − 3e−t
1
1 2t
2e − 2
!
.
!
.
[30%] Ch8(c): Check the correct statements.
1. The system u0 = Au can only be solved when A is diagonalizable.
2. The matrix exponential eAt can be found using Laplace theory.
~
3. A second order system ~x00 = A~x + G(t)
can be transformed into a first order system of the form
0
~
~u = B~u + F(t).
Answer: The first is false. The others are true.
Staple this page to the top of all Ch8 work.
Name
12:55pm Version 1
Differential Equations and Linear Algebra 2250-4 [12:55 class]Final Exam at 1:00pm on 3 May 2010
Ch9. (Nonlinear Systems) Complete all problems.
[30%] Ch9(a):
Determine whether the equilibrium u = 0 is stable or unstable. Then classify the equilibrium point
u = 0 as a saddle, center, spiral or node.
0
u =
!
5 10
−4 −7
u
Answer: The eigenvalues of A are roots of r2 + 2r + 5 = (r + 1)2 + 4 = 0, which are complex
conjugate roots −1 ± 2i. Rotation eliminates the saddle and node. Finally, the atoms e−t cos 2t,
e−t sin 2t have limit zero at t = ∞, therefore the system is stable at t = ∞. So it must be a spiral
[centers have no exponentials]. Report: stable spiral.
[30%] Ch9(b): Consider the nonlinear dynamical system
x0 = x − y 2 + y + 9,
y 0 = 2x + 2y.
An equilibrium point is x = 3, y = −3. Compute the Jacobian matrix A of the linearized system at this
equilibrium point.
Answer: The Jacobian is J(x, y) =
1 −2y + 1
2
2
!
. Then A = J(3, −3) =
1 7
2 2
!
.
[40%] Ch9(c): Consider the nonlinear dynamical system
x0 = 4x − 4y + 9 − x2 ,
y 0 = 3x − 3y.
At equilibrium point x = −3, y = −3, the Jacobian matrix is A =
10 −4
3 −3
!
.
(1) Determine the stability at t = ∞ and the phase portrait classification saddle, center, spiral or node
at u = 0 for the linear system u0 = Au.
(2) Apply a theorem to classify x = −3, y = −3 as a saddle, center, spiral or node for the nonlinear
dynamical system.
!
!
4 − 2x −4
10 −4
Answer: The Jacobian is J(x, y) =
. Then A = J(−3, −3) =
.
3 −3
3 −3
The eigenvalues of A are found from r2 − 7r − 18 = 0, giving roots a = −2, b = 9. Because
both atoms eat , ebt are real, rotation does not happen, and the classification must be a saddle
or a node. The atoms don’t have a limit at t = ±∞, therefore it is a saddle, and we report an
unstable saddle for the linear problem u0 = Au at equilibrium u = 0. Theorem 2 in 9.2 applies
to say that the same is true for the nonlinear system: unstable saddle at x = −3, y = −3.
Staple this page to the top of all Ch9 work.
Name
12:55pm Version 1
Differential Equations and Linear Algebra 2250-4 [12:55 class]Final Exam at 1:00pm on 3 May 2010
Ch10. (Laplace Transform Methods) Complete all problems.
It is assumed that you know the minimum forward Laplace integral table and the 8 basic rules for
Laplace integrals. No other tables or theory are required to solve the problems below. If you don’t know
a table entry, then leave the expression unevaluated for partial credit.
[20%] Ch10(a): Fill in the blank spaces in the Laplace tables:
f (t)
L(f (t))
f (t)
t2
1
s
1
sn
teat
1
s−a
s2
et cos 2t
s
+ b2
s2
b
+ b2
et (t + sin 2t)
(1 + 2t)2
L(f (t))
tn−1
2
, eat , cos bt, sin bt. Second table left to right: 3 ,
(n − 1)!
s
1
s−1
1
2
1
4
8
,
,
+
, +
+ .
(s − a)2 (s − 1)2 + 4 (s − 1)2 (s − 1)2 + 4 s s2 s3
Answer: First table left to right: 1,
[20%] Ch10(b): Compute L(f (t)) for f (t) = et on t ≥ 2, f (t) = 0 otherwise.
Answer: Use f (t) = et g(t), g(t) = step(t − 2), and the second shifting theorem. Then
2
e
.
L(f (t)) = L(et step(t − 2)) = e−2s L( et t→t+2 ) = e−2s L(et+2 ) = e−2s s−1
[20%] Ch10(c): Solve for f (t) in the equation L(f (t)) =
e−s
.
s−2
−as L(h(t)) = L(h(t − a) step(t − a)). Then
Answer: Use the second shifting
theorem, e
L(f (t)) = e−s L(e−2t ) = L( e−2u u=t−1 step(t − 1)) = L(e2−2t step(t − 1)). Lerch’s theorem
implies f (t) = e2−2t step(t − 1).
[20%] Ch10(d): Solve for f (t) in the equation
Answer: L((−t)f (t)) =
d
1
d2
L(f (t)) =
+
L(sin t).
ds
(s + 1)2 ds2
1 +L((−t)2 sin t)
u2 u=s+1
= L(e−t (t)+t2 sin t), giving the final answer
f (t) = −e−t − t sin t. The Laplace details use the first shifting theorem and s-differentiation
theorem.
[20%] Ch10(e): Solve by Laplace’s method for the solution x(t):
x00 (t) + x0 (t) = e−t ,
x(0) = x0 (0) = 0.
Answer: x(t) = −(1 + t)e−t + 1. The Laplace steps are: (1) L(x(t)) =
1
; (2)
s(s + 1)2
A
B
C
1
= +
+
= L(A + Be−t + Cte−t ) where by partial fractions
2
s(s + 1)
s s + 1 (s + 1)2
the answers are A = 1, B = −1, C = −1. Lerch’s theorem implies x(t) = 1 − e−t − te−t .
L(x(t)) =
Staple this page to the top of all Ch10 work.